106. Construct Binary Tree from Inorder and Postorder Traversal

/**
* Definition for a binary tree node.
* struct TreeNode {
*     int val;
*     TreeNode *left;
*     TreeNode *right;
*     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:

TreeNode* build(int i1,int j1,int i2,int j2,vector<int>& postorder, vector<int>& inorder)
{
if(i1==j1&&i2==j2)
{
TreeNode* root=new TreeNode(postorder[i1]);
return root;
}
else if(i1>j1||i2>j2)
return NULL;
else
{
TreeNode* root=new TreeNode(postorder[j1]);
int mid;
for(mid=i2;mid<=j2;mid++)
{
if(postorder[j1]==inorder[mid])
break;
}
//mid---i1
root->left=build(i1,mid-1-i2+i1,i2,mid-1,postorder, inorder);
root->right=build(i1+mid-i2,j1-1,mid+1,j2,postorder, inorder);
return root;
}
}

TreeNode* buildTree(vector<int>& inorder, vector<int>& postorder) {
return build(0,postorder.size()-1,0,inorder.size()-1,postorder,inorder);
}
};